Problem: In a class of $10$, there are $6$ students who are martial arts masters. If the teacher chooses $3$ students, what is the probability that none of the three of them are martial arts masters?
Solution: We can think about this problem as the probability of $3$ events happening. The first event is the teacher choosing one student who is not a martial arts master. The second event is the teacher choosing another student who is not a martial arts master, given that the teacher already chose someone who is not a martial arts master, and so on. The probabilty that the teacher will choose someone who is not a martial arts master is the number of students who are not martial arts masters divided by the total number of students: $\dfrac{4} {10}$ Once the teacher's chosen one student, there are only $9$ left. There's also one fewer student who is not a martial arts master, since the teacher isn't going to pick the same student twice. So, the probability that the teacher picks a second student who also is not a martial arts master is $\dfrac{3} {9}$ The probability of the teacher picking two students who are not martial arts masters must then be $\dfrac{4} {10} \cdot \dfrac{3} {9}$ We can continue using the same logic for the rest of the students the teacher picks. So, the probability of the teacher picking $3$ students such that none of them are martial arts masters is $\dfrac{4}{10}\cdot\dfrac{3}{9}\cdot\dfrac{2}{8} = \dfrac{1}{30}$